description/proof of that product of regular topological spaces is regular
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of regular topological space.
- The reader admits the proposition that any topological space is regular if and only if for each point of the space, the 1-point subset is closed and the set of the closed neighborhoods of the point is a neighborhoods basis at the point.
- The reader admits the proposition that the product of any possibly uncountable number of Hausdorff topological spaces is Hausdorff.
- The reader admits the proposition that for any Hausdorff topological space, any 1 point subset is closed.
- The reader admits the proposition that for any product topological space and any neighborhood of any point, there is an open neighborhood of the point contained in the neighborhood as the product of some open neighborhoods of the components of the point.
- The reader admits the proposition that any subset of any product topological space is closed if and only if it is the intersection of any possibly uncountable number of finite unions of the products of any closed subsets only finite number of which (the subsets) are not the whole spaces, and especially only if only 1 of the subsets is not the whole space.
Target Context
- The reader will have a description and a proof of the proposition that the product of any possibly uncountable number of regular topological spaces is regular.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \in \{\text{ the regular topological spaces }\} \vert j \in J\}\):
\(\times_{j \in J} T_j\): \(= \text{ the product topological space }\)
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Statements:
\(\times_{j \in J} T_j \in \{\text{ the regular topological spaces }\}\)
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2: Proof
Whole Strategy: apply the proposition that any topological space is regular if and only if for each point of the space, the 1-point subset is closed and the set of the closed neighborhoods of the point is a neighborhoods basis at the point; Step 1: see that for each \(t \in \times_{j \in J} T_j\), \(\{t\}\) is a closed subset; Step 2: see that for each \(t \in \times_{j \in J} T_j\), the set of the closed neighborhoods of \(t\) is a neighborhoods basis at \(t\); Step 3: conclude the proposition.
Step 1:
For each \(j \in J\), \(T_j\) is Hausdorff, because for each \(t_j, t'_j \in T_j\) such that \(t_j \neq t'_j\), \(\{t'_j\}\) is a closed subset, and there are an open neighborhood of \(t_j\), \(U_{t_j} \subseteq T_j\), and an open neighborhood of \(\{t'_j\}\), which is an open neighborhood of \(t'_j\), \(U_{t'_j} \subseteq T_j\), such that \(U_{t_j} \cap U_{t'_j} = \emptyset\).
So, \(\times_{j \in J} T_j\) is Hausdorff, by the proposition that the product of any possibly uncountable number of Hausdorff topological spaces is Hausdorff.
For each \(t \in \times_{j \in J} T_j\), \(\{t\}\) is closed, by the proposition that for any Hausdorff topological space, any 1 point subset is closed.
Step 2:
Let us see that for each \(t \in \times_{j \in J} T_j\), the set of the closed neighborhoods of \(t\), \(S_t\), is a neighborhoods basis at \(t\).
Let \(N_t \subseteq \times_{j \in J} T_j\) be any neighborhood of \(t\).
There are a finite \(J^` \subseteq J\) and \(\{U_{t^j} \in \{\text{ the open neighborhoods of } t^j \text{ on } T_j\} \vert j \in J^`\}\) such that \(\times_{j \in J} U_{t^j}\), where \(U_{t^j} := T_j\) for each \(j \in J \setminus J^`\), is an open neighborhood of \(t\) such that \(\times_{j \in J} U_{t^j} \subseteq N_t\), by the proposition that for any product topological space and any neighborhood of any point, there is an open neighborhood of the point contained in the neighborhood as the product of some open neighborhoods of the components of the point.
For each \(j \in J^`\), there is a closed neighborhood of \(t^j\), \(C_{t^j} \subset T_j\), such that \(C_{t^j} \subseteq U_{t^j}\), by the proposition that any topological space is regular if and only if for each point of the space, the 1-point subset is closed and the set of the closed neighborhoods of the point is a neighborhoods basis at the point, because \(T_j\) is regular.
For each \(j \in J \setminus J^`\), take the closed neighborhood of \(t^j\), \(C_{t^j} := T_j\), which satisfies \(C_{t^j} = T_j \subseteq T_j = U_{t^j}\).
Let us see that \(\times_{j \in J} C_{t^j} \subseteq \times_{j \in J} T_j\) is a closed neighborhood of \(t\).
\(t \in \times_{j \in J} C_{t^j}\).
\(\times_{j \in J} C_{t^j}\) is closed, by the proposition that any subset of any product topological space is closed if and only if it is the intersection of any possibly uncountable number of finite unions of the products of any closed subsets only finite number of which (the subsets) are not the whole spaces, and especially only if only 1 of the subsets is not the whole space.
For each \(j \in J^`\), there is an open neighborhood of \(t^j\), \(V_{t^j} \subseteq C_{t^j}\), because \(C_{t^j}\) is a neighborhood of \(t^j\).
For each \(j \in J \setminus J^`\), let us take the open neighborhood of \(t^j\), \(V_{t^j} := T_j\), which satisfies \(V_{t^j} = T_j \subseteq T_j = C_{t^j}\).
Then, \(t \in \times_{j \in J} V_{t^j} \subseteq \times_{j \in J} C_{t^j}\), where \(\times_{j \in J} V_{t^j} \subseteq \times_{j \in J} T_j\) is open.
So, \(\times_{j \in J} C_{t^j}\) is a closed neighborhood of \(t\).
So, \(\times_{j \in J} C_{t^j} \in S_t\).
\(\times_{j \in J} C_{t^j} \subseteq \times_{j \in J} U_{t^j} \subseteq N_t\).
So, \(S_t\) is a neighborhoods basis at \(t\).
Step 3:
So, by the proposition that any topological space is regular if and only if for each point of the space, the 1-point subset is closed and the set of the closed neighborhoods of the point is a neighborhoods basis at the point, \(\times_{j \in J} T_j\) is regular.
